# Evaluate the integral int \ 1/(x^2sqrt(x^2-9)) \ dx ?

Oct 18, 2017

$\frac{\sqrt{{x}^{2} - 9}}{9 | x |} + C .$

#### Explanation:

We substitute, $x = 3 \sec y \Rightarrow \mathrm{dx} = 3 \sec y \tan y \mathrm{dy} .$

$\therefore I = \int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 9}} \mathrm{dx} ,$

=int{(3secy tany)dy}/{9sec^2ysqrt(9sec^2y-9),

$= \int \frac{\left(\cancel{3} \sec y \cancel{\tan y}\right) \mathrm{dy}}{9 {\sec}^{2} y \cdot \cancel{3 \tan y}} ,$

$= \frac{1}{9} \int \frac{1}{\sec} y \mathrm{dy} = \frac{1}{9} \int \cos y \mathrm{dy} ,$

$= \frac{1}{9} \sin y ,$

$= \frac{1}{9} \sqrt{1 - {\cos}^{2} y} ,$

$= \frac{1}{9} \sqrt{1 - \frac{1}{\sec} ^ 2 y} ,$

$= \frac{1}{9} \sqrt{1 - {\left(\frac{3}{x}\right)}^{2}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left[\because , x = 3 \sec y\right] ,$

$= \frac{1}{9} \frac{\sqrt{{x}^{2} - 9}}{|} x | .$

$\Rightarrow I = \frac{\sqrt{{x}^{2} - 9}}{9 | x |} + C .$

Enjoy Maths.!

Oct 18, 2017

$\int \setminus \frac{1}{{x}^{2} \sqrt{{x}^{2} - 9}} \setminus \mathrm{dx} = \frac{1}{9} \sqrt{1 - \frac{9}{x} ^ 2} + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{{x}^{2} \sqrt{{x}^{2} - 9}} \setminus \mathrm{dx}$

We can write the integral as follows:

$I = \int \setminus \frac{1}{{x}^{2} \sqrt{9 \left({x}^{2} / 9 - 1\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{{x}^{2} \sqrt{9} \sqrt{{\left(\frac{x}{3}\right)}^{2} - 1}} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{3} \setminus \int \setminus \frac{1}{{x}^{2} \sqrt{{\left(\frac{x}{3}\right)}^{2} - 1}} \setminus \mathrm{dx}$

Let us attempt a substitution of the form:

$3 \sec \theta = x \implies 3 \sec \theta \tan \theta \frac{d \theta}{\mathrm{dx}} = 1$

Then substituting into the integral, we get:

$I = \frac{1}{3} \setminus \int \setminus \frac{1}{{\left(3 \sec \theta\right)}^{2} \sqrt{{\left(\sec \theta\right)}^{2} - 1}} \setminus 3 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus = \int \setminus \frac{\sec \theta \tan \theta}{9 {\sec}^{2} \theta \sqrt{{\sec}^{2} \theta - 1}} \setminus d \theta$

$\setminus \setminus = \frac{1}{9} \setminus \int \setminus \frac{\tan \theta}{\sec \theta \sqrt{{\tan}^{2} \theta}} \setminus d \theta$

$\setminus \setminus = \frac{1}{9} \setminus \int \setminus \frac{\tan \theta}{\sec \theta \tan \theta} \setminus d \theta$

$\setminus \setminus = \frac{1}{9} \setminus \int \setminus \frac{1}{\sec \theta} \setminus d \theta$

$\setminus \setminus = \frac{1}{9} \setminus \int \setminus \cos \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{9} \sin \theta + C$

$\setminus \setminus = \frac{1}{9} \sqrt{1 - {\cos}^{2} \theta} + C$

$\setminus \setminus = \frac{1}{9} \sqrt{1 - \left(\frac{1}{\sec} ^ 2 \theta\right)} + C$

And we can now restore the earlier substitution:

$I = \frac{1}{9} \sqrt{1 - \left(\frac{1}{\frac{x}{3}} ^ 2\right)} + C$
 \ \ = 1/9 sqrt(1-(1/(x^2/9)) + C
$\setminus \setminus = \frac{1}{9} \sqrt{1 - \frac{9}{x} ^ 2} + C$

Oct 18, 2017

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{\sqrt{{x}^{2} - 9}}{9 x} + C$

#### Explanation:

Restricting to the interval $x \in \left(3 , + \infty\right)$, substitute:

$x = 3 \sec t$

$\mathrm{dx} = 3 \sec t \tan t \mathrm{dt}$

with $t \in \left(0 , \frac{\pi}{2}\right)$ and get:

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = 3 \int \frac{\sec t \tan t \mathrm{dt}}{9 {\sec}^{2} t \sqrt{9 {\sec}^{2} t - 9}}$

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \int \frac{\tan t \mathrm{dt}}{\sec t \sqrt{{\sec}^{2} t - 1}}$

use now the trigonometric identity:

${\sec}^{2} t - 1 = \frac{1}{\cos} ^ 2 - 1 = \frac{1 - {\cos}^{2} t}{\cos} ^ 2 t = {\sin}^{2} \frac{t}{\cos} ^ 2 t = {\tan}^{2} t$

so that:

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \int \frac{\tan t \mathrm{dt}}{\sec t \sqrt{{\tan}^{2} t}}$

and as in the selected interval the tangent is positive:

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \int \frac{\tan t \mathrm{dt}}{\sec t \tan t}$

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \int \frac{\mathrm{dt}}{\sec} t$

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \int \cos t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{1}{9} \sin t + C$

To undo the substitution note that:

$x = 3 \sec t = \frac{3}{\cos} t = \frac{3}{\sqrt{1 - {\sin}^{2} t}}$

$\sqrt{1 - {\sin}^{2} t} = \frac{3}{x}$

$1 - {\sin}^{2} t = \frac{9}{x} ^ 2$

${\sin}^{2} t = 1 - \frac{9}{x} ^ 2$

and as in the interval the sine is positive:

$\sin t = \frac{\sqrt{{x}^{2} - 9}}{x}$

So:

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}} = \frac{\sqrt{{x}^{2} - 9}}{9 x} + C$

And using the substitution $x = - \sec t$ for $x \in \left(- \infty , - 3\right)$ we obtain the same result.